YES
* Step 1: UnsatPaths YES
    + Considered Problem:
        Rules:
          0. start(A,B,C,D,E,F,G,H)  -> a(A,B,C,D,E,F,G,H)      [A >= 1 && B = A && C = D && E = F && G = H]                                (?,1)
          1. a(A,B,C,D,E,F,G,H)      -> d(A,B,1,D,E,F,G,H)      [A >= 1 && G = H && E = F && C = D && B = A]                                (?,1)
          2. d(A,B,C,D,E,F,G,H)      -> b(A,B,C,D,E,F,1 + C,H)  [A >= 1 + C && A >= C && A >= 1 && C >= 1 && B = A]                         (?,1)
          3. d(A,B,C,D,E,F,G,H)      -> halt(A,B,C,D,E,F,G,H)   [A >= 1 && C = A && B = A]                                                  (?,1)
          4. b(A,B,C,D,E,F,G,H)      -> d(A,B,1 + C,D,E,F,G,H)  [A >= 1 + C && C >= 1 && A >= C && G = 1 + A && B = A]                      (?,1)
          5. b(A,B,C,D,E,F,G,H)      -> c(A,B,C,D,B,F,G,H)      [A >= G && 1 + A >= G && A >= 1 + C && C >= 1 && G >= 1 + C && B = A]       (?,1)
          6. c(A,B,C,D,E,F,G,H)      -> c(A,B,C,D,-1 + E,F,G,H) [E >= 1 + C && A >= E && A >= G && C >= 1 && E >= C && G >= 1 + C && B = A] (?,1)
          7. c(A,B,C,D,E,F,G,H)      -> b(A,B,C,D,E,F,1 + G,H)  [A >= E && A >= G && E >= 1 && G >= 1 + E && C = E && B = A]                (?,1)
          8. start0(A,B,C,D,E,F,G,H) -> start(A,A,D,D,F,F,H,H)  True                                                                        (1,1)
        Signature:
          {(a,8);(b,8);(c,8);(d,8);(halt,8);(start,8);(start0,8)}
        Flow Graph:
          [0->{1},1->{2,3},2->{4,5},3->{},4->{2,3},5->{6,7},6->{6,7},7->{4,5},8->{0}]
        
    + Applied Processor:
        UnsatPaths
    + Details:
        We remove following edges from the transition graph: [(2,4),(5,7)]
* Step 2: FromIts YES
    + Considered Problem:
        Rules:
          0. start(A,B,C,D,E,F,G,H)  -> a(A,B,C,D,E,F,G,H)      [A >= 1 && B = A && C = D && E = F && G = H]                                (?,1)
          1. a(A,B,C,D,E,F,G,H)      -> d(A,B,1,D,E,F,G,H)      [A >= 1 && G = H && E = F && C = D && B = A]                                (?,1)
          2. d(A,B,C,D,E,F,G,H)      -> b(A,B,C,D,E,F,1 + C,H)  [A >= 1 + C && A >= C && A >= 1 && C >= 1 && B = A]                         (?,1)
          3. d(A,B,C,D,E,F,G,H)      -> halt(A,B,C,D,E,F,G,H)   [A >= 1 && C = A && B = A]                                                  (?,1)
          4. b(A,B,C,D,E,F,G,H)      -> d(A,B,1 + C,D,E,F,G,H)  [A >= 1 + C && C >= 1 && A >= C && G = 1 + A && B = A]                      (?,1)
          5. b(A,B,C,D,E,F,G,H)      -> c(A,B,C,D,B,F,G,H)      [A >= G && 1 + A >= G && A >= 1 + C && C >= 1 && G >= 1 + C && B = A]       (?,1)
          6. c(A,B,C,D,E,F,G,H)      -> c(A,B,C,D,-1 + E,F,G,H) [E >= 1 + C && A >= E && A >= G && C >= 1 && E >= C && G >= 1 + C && B = A] (?,1)
          7. c(A,B,C,D,E,F,G,H)      -> b(A,B,C,D,E,F,1 + G,H)  [A >= E && A >= G && E >= 1 && G >= 1 + E && C = E && B = A]                (?,1)
          8. start0(A,B,C,D,E,F,G,H) -> start(A,A,D,D,F,F,H,H)  True                                                                        (1,1)
        Signature:
          {(a,8);(b,8);(c,8);(d,8);(halt,8);(start,8);(start0,8)}
        Flow Graph:
          [0->{1},1->{2,3},2->{5},3->{},4->{2,3},5->{6},6->{6,7},7->{4,5},8->{0}]
        
    + Applied Processor:
        FromIts
    + Details:
        ()
* Step 3: Decompose YES
    + Considered Problem:
        Rules:
          start(A,B,C,D,E,F,G,H)  -> a(A,B,C,D,E,F,G,H)       [A >= 1 && B = A && C = D && E = F && G = H]
          a(A,B,C,D,E,F,G,H)      -> d(A,B,1,D,E,F,G,H)       [A >= 1 && G = H && E = F && C = D && B = A]
          d(A,B,C,D,E,F,G,H)      -> b(A,B,C,D,E,F,1 + C,H)   [A >= 1 + C && A >= C && A >= 1 && C >= 1 && B = A]
          d(A,B,C,D,E,F,G,H)      -> halt(A,B,C,D,E,F,G,H)    [A >= 1 && C = A && B = A]
          b(A,B,C,D,E,F,G,H)      -> d(A,B,1 + C,D,E,F,G,H)   [A >= 1 + C && C >= 1 && A >= C && G = 1 + A && B = A]
          b(A,B,C,D,E,F,G,H)      -> c(A,B,C,D,B,F,G,H)       [A >= G
                                                               && 1 + A >= G
                                                               && A >= 1 + C
                                                               && C >= 1
                                                               && G >= 1 + C
                                                               && B = A]
          c(A,B,C,D,E,F,G,H)      -> c(A,B,C,D,-1 + E,F,G,H)  [E >= 1 + C
                                                               && A >= E
                                                               && A >= G
                                                               && C >= 1
                                                               && E >= C
                                                               && G >= 1 + C
                                                               && B = A]
          c(A,B,C,D,E,F,G,H)      -> b(A,B,C,D,E,F,1 + G,H)   [A >= E
                                                               && A >= G
                                                               && E >= 1
                                                               && G >= 1 + E
                                                               && C = E
                                                               && B = A]
          start0(A,B,C,D,E,F,G,H) -> start(A,A,D,D,F,F,H,H)   True
        Signature:
          {(a,8);(b,8);(c,8);(d,8);(halt,8);(start,8);(start0,8)}
        Rule Graph:
          [0->{1},1->{2,3},2->{5},3->{},4->{2,3},5->{6},6->{6,7},7->{4,5},8->{0}]
    + Applied Processor:
        Decompose NoGreedy
    + Details:
        We construct a looptree:
          P: [0,1,2,3,4,5,6,7,8]
          |
          `- p:[2,4,7,6,5] c: [2,4]
             |
             `- p:[5,7,6] c: [5,7]
                |
                `- p:[6] c: [6]
* Step 4: CloseWith YES
    + Considered Problem:
        (Rules:
          start(A,B,C,D,E,F,G,H)  -> a(A,B,C,D,E,F,G,H)       [A >= 1 && B = A && C = D && E = F && G = H]
          a(A,B,C,D,E,F,G,H)      -> d(A,B,1,D,E,F,G,H)       [A >= 1 && G = H && E = F && C = D && B = A]
          d(A,B,C,D,E,F,G,H)      -> b(A,B,C,D,E,F,1 + C,H)   [A >= 1 + C && A >= C && A >= 1 && C >= 1 && B = A]
          d(A,B,C,D,E,F,G,H)      -> halt(A,B,C,D,E,F,G,H)    [A >= 1 && C = A && B = A]
          b(A,B,C,D,E,F,G,H)      -> d(A,B,1 + C,D,E,F,G,H)   [A >= 1 + C && C >= 1 && A >= C && G = 1 + A && B = A]
          b(A,B,C,D,E,F,G,H)      -> c(A,B,C,D,B,F,G,H)       [A >= G
                                                               && 1 + A >= G
                                                               && A >= 1 + C
                                                               && C >= 1
                                                               && G >= 1 + C
                                                               && B = A]
          c(A,B,C,D,E,F,G,H)      -> c(A,B,C,D,-1 + E,F,G,H)  [E >= 1 + C
                                                               && A >= E
                                                               && A >= G
                                                               && C >= 1
                                                               && E >= C
                                                               && G >= 1 + C
                                                               && B = A]
          c(A,B,C,D,E,F,G,H)      -> b(A,B,C,D,E,F,1 + G,H)   [A >= E
                                                               && A >= G
                                                               && E >= 1
                                                               && G >= 1 + E
                                                               && C = E
                                                               && B = A]
          start0(A,B,C,D,E,F,G,H) -> start(A,A,D,D,F,F,H,H)   True
        Signature:
          {(a,8);(b,8);(c,8);(d,8);(halt,8);(start,8);(start0,8)}
        Rule Graph:
          [0->{1},1->{2,3},2->{5},3->{},4->{2,3},5->{6},6->{6,7},7->{4,5},8->{0}]
        ,We construct a looptree:
          P: [0,1,2,3,4,5,6,7,8]
          |
          `- p:[2,4,7,6,5] c: [2,4]
             |
             `- p:[5,7,6] c: [5,7]
                |
                `- p:[6] c: [6])
    + Applied Processor:
        CloseWith True
    + Details:
        ()

YES